An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its re

Question

An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is $\_\_\_\_$ . $(\pi=3.14)$

Professor PC Thomas Classes · Answer

The average thermal power dissipated by the resistive component is $P = I_{rms}^2 R \implies 32 = (2)^2 R \implies 4R = 32 \implies R = 8 \Omega$. The average magnetic energy stored in an inductor carrying an AC current is $U_{avg} = \frac{1}{2} L I_{rms}^2$. Using the given $16 	ext{ J}$, $16 = \frac{1}{2} L (2)^2 \implies 16 = 2L \implies L = 8 	ext{ H}$. The inductive reactance is $X_L = \omega L = 2\pi f L = 2 	imes 3.14 	imes 50 	imes 8 = 314 	imes 8 = 2512 \Omega$. The ratio of inductive reactance to resistance is $\frac{X_L}{R} = \frac{2512}{8} = 314$.